Hibbeler Statics 14E P2.1 — Solving for the Magnitude and Direction of the Resultant of Two Coplanar-Concurrent Forces

If θ=60°\theta = 60 \degree and F=450 N \textbf{F} = 450 \ \text{N}, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-1

Solution:

The parallelogram law and the triangulation rule are shown in the figures below.

(a) Parallelogram Law
(b) Triangulation Rule

Considering figure (b), we can solve for the magnitude of FR \textbf{F}_R using the cosine law.

FR=7002+4502−2(700)(450)cos⁡45∘=497.01 N=497 N\begin{align*} \textbf{F}_R & = \sqrt{700^2+450^2-2\left( 700 \right)\left( 450 \right)\cos45^{\circ}}\\ & = 497.01 \ \text{N}\\ & = 497 \ \text{N} \end{align*}

Then we use the sine law to solve for the interior angle θ\theta.

sin⁡θ700=sin⁡45∘497.01sin⁡θ=700 sin⁡45∘497.01θ=sin⁡−1(700 sin⁡45∘497.01)This is an ambiguous case θ=84.81∘ or θ=95.19∘\begin{align*} \frac{\sin \theta}{700} & = \frac{\sin 45^{\circ}}{497.01}\\ \sin \theta & =\frac{700\ \sin 45^{\circ }}{497.01}\\ \theta & = \sin^{-1} \left( \frac{700\ \sin 45^{\circ }}{497.01} \right)\\ & \text{This is an ambiguous case }\\ \theta & = 84.81^\circ \ or \ \theta =95.19^\circ \\ \end{align*}

In here, the correct angle measurement is θ=95.19∘ \theta = 95.19^{\circ}.

Thus, the direction angle ϕ \phi of FR \textbf{F}_R measured counterclockwise from the positive x-axis, is

ϕ=θ+60∘=95.19∘+60∘=155∘\begin{align*} \phi & = \theta +60^\circ \\ & = 95.19^\circ +60^\circ \\ & = 155^\circ \end{align*}
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